title: How to create Cube with inner shadow to corner with curve?
body: <p>I am trying to create this image with css and html but m kept failing to get perfect curve and shadow as shown in image.</p>
  
  <p><a href="https://i.stack.imgur.com/0Y3vw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0Y3vw.png" alt="Image"></a></p>
  
  <p><strong>HTML</strong></p>
  
  <pre><code>&lt;div class="mainDiv"&gt;
    &lt;div class="square"&gt;&lt;/div&gt;
    &lt;div class="square2"&gt;&lt;/div&gt;
    &lt;div class="square3"&gt;&lt;/div&gt;
  &lt;/div&gt;
  </code></pre>
  
  <p><strong>CSS</strong></p>
  
  <pre><code>.mainDiv{
    position: relative;
    width: 206px;
    height: 190px;
    margin: 0px auto;
    margin-top:100px;
  }
  .square{
    width:100px;
    height:100px;
    background:#df1c40;
    float:left;
    transform: skew(180deg,210deg);
    position: absolute;
    top: 42px;
  }
  .square2{
    width:100px;
    height:100px;
    background:#ff9c28;
    float:left;
    transform: skew(180deg,150deg);
    position: absolute;
    left:100px;
    top: 42px;
  }
  .square3{
    width:117px;
    height:100px;
    background:#97bbdd;
    float:left;
    transform: rotate(150deg) translate(-40px, -16px) skew(30deg, 0deg);
    position: absolute;
    left: -1px;
    top: -31px;
  }
  .square3:before {
    content: "";
    position: absolute;
    top: 0%;
    right: 0%;
    width: 0px;
    height: 0px;
    border-bottom: 70px solid #28a55c;
    border-left: 70px solid transparent;
    -webkit-box-shadow: 5px 0px 0px -1px #879fd1;
    -moz-box-shadow: 5px 0px 0px -1px #879fd1;
    box-shadow: 5px 0px 0px -1px #879fd1;
    transform: rotate(90deg);
  }
  .square3:after {
    content: "";
    position: absolute;
    top: 0%;
    right: 0%;
    width: 0px;
    height: 0px;
    border-top: 70px solid #F3F5F6; 
    border-right: 70px solid transparent;
    transform: rotate(90deg);
  }
  </code></pre>
  
  <p>My best <a href="https://jsfiddle.net/LeoLion/8vuj7peb/64/" rel="nofollow noreferrer">try</a>. I am just doing this for my knowledge and any help with little bit detail will be appreciate.. :)</p>